Stirling permutations, marked permutations and Stirling derangements

STIRLING PERMUTATIONS, MARKED PERMUTATIONS AND STIRLING DERANGEMENTS

arXiv:1612.07391v1 [math.CO] 21 Dec 2016

GUAN-HUEI DUH, YEN-CHI ROGER LIN, SHI-MEI MA, AND YEONG-NAN YEH Abstract. In this paper we introduce the definition of marked permutations. We first present a bijection between Stirling permutations and marked permutations. We then present an involution on Stirling derangements. Furthermore, we present a symmetric bivariate enumerative polynomials on r-colored marked permutations. Finally, we give an explanation of r-colored marked permutations by using the language of combinatorial objects.

1. Introduction Let Sn denote the symmetric group of all permutations of [n], where [n] = {1, 2, . . . , n}. Let π = π1 π2 · · · πn ∈ Sn . An ascent of π is an entry πi , i ∈ {2, 3, . . . , n}, such that πi > πi−1 . Denote by ASC(π) the set of all ascents of π, and asc(π) = |ASC(π)|. For example, asc(53412) = |{2, 4}| = 2. The classical Eulerian polynomial is defined by X xasc(π) . An (x) = π∈Sn

Set A0 (x) = 1. The exponential generating function for An (x) is X 1−x tn . A(x, t) = An (x) = t(x−1) n! e −x

(1)

n≥0

Stirling permutations were introduced by Gessel and Stanley [7]. Let [n]2 denote the multiset {1, 1, 2, 2, . . . , n, n}. A Stirling permutation of order n is a permutation of [n]2 such that every entry between the two occurrences of i is greater than i for each i ∈ [n]. Various statistics on Stirling permutations have been extensively studied in the past decades, including descents [3, 7, 8], plateaux [1, 3, 8], blocks [13], ascent plateaux [10, 11] and cycle ascent plateaux [12]. Denote by Qn the set of Stirling permutations of order n and let σ = σ1 σ2 · · · σ2n ∈ Qn . An occurrence of an ascent plateau is an entry σi , i ∈ {2, 3, . . . , 2n − 1}, such that σi−1 < σi = σi+1 (see [10]). Let AP(σ) be the set of all ascent plateaux of σ, and ap(σ) = |AP(σ)|. As an example, ap(221133) = |{3}| = 1. Let X xap(π) . Nn (x) = π∈Qn

Set N0 (x) = 1. From [10, Theorem 2], we have X tn p Nn (x) = A(x, 2t). n!

(2)

n≥0

2010 Mathematics Subject Classification. Primary 05A15; Secondary 26C10. Key words and phrases. Stirling permutations; Marked permutations; Stirling derangements; Increasing trees. 1

2

G.-H. DUH, Y.-C. LIN, S.-M. MA, AND Y.-N. YEH

Let w = w1 w2 · · · wn be a word on [n]. A left-to-right minimum of w is an element wi such that wi < wj for every j ∈ [i−1] or i = 1; a right-to-left minimum of w is an element wi such that wi < wj for every j ∈ {i + 1, i + 2, . . . , n} or i = n. Let LRMIN(w) and RLMIN(w) denote the set of entries of left-to-right minima and right-to-left minima of w, respectively. Set lrmin(w) = |LRMIN(w)| and rlmin(w) = |RLMIN(w)|. For example, lrmin(223311) = |{1, 2}| = 2 and rlmin(223311) = |{1}| = 1. Motivated by (2), we now introduce the definition of marked permutations. Given a permutation π = π1 π2 · · · πn ∈ Sn , a marked permutation is a permutation with marks on some of its non-left-to-right minima. An element i is denoted by ı when it is marked. Let Sn denote the set of marked permutations of [n]. For example, S1 = {1}, S2 = {12, 12, 21} and S3 ={1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, 3 2 1, 1 2 3, 1 2 3, 1 2 3, 1 3 2, 1 3 2, 1 3 2, 2 1 3, 2 3 1, 3 1 2}. This paper is organized as follows. In Section 2, we present a bijection between Stirling permutations and marked permutations. In Section 3, we use an involution to prove an identity between Stirling derangements and perfect matchings. In Section 4, we present a symmetric bivariate enumerative polynomials on r-colored marked permutations. And in Section 5, we use the language of combinatorial objects to give an explanation of marked permutations. 2. A bijection between Stirling permutations and marked permutations Let σ = σ1 σ2 · · · σ2n ∈ Qn be a Stirling permutation of order n. An entry k of σ is called an even indexed entry (resp. odd indexed entry) if the first appearance of k occurs at an even (resp. odd) position of σ. Let EVEN(σ) (resp. ODD(σ)) denote the set of even (resp. odd) indexed entries of σ, even(σ) = |EVEN(σ)|, and odd(σ) = |ODD(σ)|. For example, even(221331) = |{3}| = 1 and odd(221331) = |{1, 2}| = 2. Given π ∈ Sn , let MARK(π) denote the set of marked entries of π, and mark(π) = |MARK(π)|. The statistics ASC(π), asc(π), LRMIN(π) and lrmin(π) are defined by forgetting the marks of π. For example, asc(3521467) = 4, mark(3521467) = 2 and lrmin(3521467) = 3. A block in an element of Qn or Sn is a substring which begins with a left-to-right minimum, and contains exactly this one left-to-right minimum; moreover, the substring is maximal, i.e. not contained in any larger such substring. It is easily derived by induction that any Stirling permutation or permutation has a unique decomposition as a sequence of blocks. Example 1. The block decompositions of 34664325527711 and 3462571 are respectively given by [346643][255277][11] and [346][257][1]. We now present the first result of this paper. Theorem 2. For n ≥ 1, we have X X q lrmin(π) xasc(π) y mark(π) . q lrmin(σ) xap(σ) y even(σ) = σ∈Qn

π∈Sn

(3)


3

Proof. We prove a stronger result: there is a one-to-one correspondence between the set statistics (LRMIN, AP, EVEN) on Stirling permutations and (LRMIN, ASC, MARK) on marked permutations. Define Q(n; X, Y, Z) = {σ ∈ Qn | LRMIN(σ) = X, AP(σ) = Y, EVEN(σ) = Z}, and S(n; X, Y, Z) = {π ∈ Sn | LRMIN(π) = X, ASC(π) = Y, MARK(π) = Z}, where X, Y, Z ⊆ [n]. Clearly, they form partitions of all Stirling permutations and marked permutations respectively. Now we start to construct a bijection, denoted by Φ, between Stirling permutations and marked permutations. In addition, it maps each Q(n; X, Y, Z) onto S(n; X, Y, Z). When n = 1, it is clear that 11 ∈ Q(1; {1}, ∅, ∅) and 1 ∈ S(1; {1}, ∅, ∅). Hence setting Φ(11) = 1 satisfies the requirement. Fix m ≥ 2, and assume that Φ is a bijection between Q(m − 1; X, Y, Z) and S(m − 1; X, Y, Z) for all possible X, Y, Z. Let σ ′ ∈ Qm be obtained from some σ ∈ Q(m − 1; X, Y, Z) by inserting the substring mm into σ. By the assumption, Φ(σ) = π ∈ S(m − 1; X, Y, Z). If mm is placed at the front of σ, that is, σ ′ = mmσ, then we let Φ(σ ′ ) = mπ. In this case, we have σ ′ ∈ Q(m; X ∪ {m}, Y, Z) and Φ(σ ′ ) ∈ S(m; X ∪ {m}, Y, Z). Notice that this is the only way to produce a new block. Otherwise, suppose σ ′ is obtained from σ by inserting mm into the pth block. Let r be the left-to-right minimum contained in the pth block of σ. There are three possible cases: (i) If mm is inserted immediately before the second r, then Φ(σ ′ ) = π ′ is obtained by inserting a marked m at the end of the pth block of π. Note that m is an additional even indexed entry, as well an ascent plateau, of σ ′ after inserting mm into σ. Meanwhile, m is a marked element, as well an ascent, of π ′ . Hence σ ′ ∈ Q(m; X, Y ∪ {m}, Z ∪ {m}) and Φ(σ ′ ) ∈ S(m; X, Y ∪ {m}, Z ∪ {m}). (ii) If mm is inserted immediately before s, s 6= r, then Φ(σ ′ ) = π ′ is obtained by inserting m or m into the pth block of π such that m is immediately before s. The inserted entry m is marked if and only if m is an even indexed entry of σ ′ . When s ∈ Y , let Y ′ = (Y ∪ {m}) \{s}. Otherwise, let Y ′ = Y ∪ {m}. When m is an even indexed entry, let Z ′ = Z ∪ {m}. Otherwise, let Z ′ = Z. In each possible case, we see that σ ′ ∈ Q(m; X, Y ′ , Z ′ ) and Φ(σ ′ ) ∈ S(m; X, Y ′ , Z ′ ). (iii) If mm is inserted at the end of the pth block, then Φ(σ ′ ) = π ′ is obtained by inserting an unmarked m at the end of the pth block of π. Note that σ does not gain any additional even indexed entry after inserting m, but obtain the ascent plateau of m. On the other hand, m is a new ascent of π ′ after inserting m into π. Hence σ ′ ∈ Q(m; X, Y ∪ {m}, Z) and Φ(σ ′ ) ∈ S(m; X, Y ∪ {m}, Z). The above argument shows that Φ(Qn ) ⊆ Sn , and that Φ is one-to-one on Qn . Since the cardinality of Qn is the same as that of Sn , Φ must be a bijection between Qn and Sn . By induction, we see that Φ is the desired bijection between Stirling permutations and marked permutations.

4


Example 3. Consider σ = 266255133441 ∈ Q6 . The correspondence between σ and Φ(σ) is built up as follows: [11] ⇔ [1] [22][11] ⇔ [2] [1] [22][1331] ⇔ [2] [1 3] [22][134431] ⇔ [2] [1 4 3] [2255][134431] ⇔ [2 5] [1 4 3] [266255][134431] ⇔ [2 5 6] [1 4 3] Exploiting the bijection Φ used in the proof of Theorem 2, we also get the following result. Theorem 4. For n ≥ 1, we have X X xap(σ) (−1)even(σ) = xasc(π) (−1)mark(π) = 1. σ∈Qn

(4)

π∈Sn

Proof. From Theorem 2, we get the first equality of (4). For the last equality of (4), we will consider an involution ι on Sn as follows. For a marked permutation π that has non-leftto-right minima, define ι(π) to be the marked permutation obtained from π by changing the marking of the smallest non-left-to-right minimum of π. For example, when π = 2 5 3 1 4, then ι(π) = 2 5 3 1 4. This map ι is clearly a sign-reversing involution because the number of marks is either plus 1 or minus 1. It is also clear that ι preserves the ascent statistic since no entry leaves its position. The only marked permutation in Sn which cannot be mapped by ι is the one in which every entry is a left-to-right minimum, i.e., the marked permutation n (n − 1) · · · 2 1, whose ascent is 0. This completes the proof. Theorem 5. We have X X

n≥0 π∈Sn

q lrmin(π) xasc(π) y mark(π)

q tn = A x, t(1 + y) 1+y , n!

where A(x, z) is the exponential generating function given by (1). Proof. Combining [2, Proposition 7.3] and the fundamental transformation introduced by Foata and Sch¨ utzenberger [6], we have X X tn X X cyc(π) exc(π) tn q lrmin(π) xasc(π) = = A(x, t)q q x n! n! n≥0 π∈Sn

n≥0 π∈Sn

For a permutation π in Sn with lrmin(π) = ℓ, there are n − ℓ entries that could be either marked or not. Therefore, we have X X tn tn X X lrmin(π) asc(π) q x (1 + y)n−lrmin(π) q lrmin(π) xasc(π) y mark(π) = n! n! n≥0 π∈Sn

n≥0 π∈Sn

X X q lrmin(π) (t(1 + y))n = xasc(π) 1+y n! n≥0 π∈Sn

q = A x, t(1 + y) 1+y .


5

Combining Theorem 2 and Theorem 5, we have X X 1 tn X X mark(π) tn y = A 1, t(1 + y) 1+y . y even(σ) = n! n! n≥0 π∈Sn

n≥0 σ∈Qn

n

Let i be the (signless) Stirling number of the first kind, i.e., the number of permutations of Sn with i cycles. Note that even(σ) + odd(σ) = n for σ ∈ Qn . Let X podd(σ) q even(σ) . En (p, q) = σ∈Qn

Now we present the following result. Theorem 6. For n ≥ 1, we have En (p, q) =

n X n k=0

k

pk (p + q)n−k .

(5)

In particular, En (1, 1) = (2n − 1)!!, En (p, 0) = n!pn , En (1, −1) = 1, En (−1, 1) = (−1)n for n ≥ 2. Proof. There are two ways in which a permutation σ ′ ∈ Qn with even(σ ′ ) = k can be obtained from a permutation σ = σ1 σ2 · · · σ2n−2 ∈ Qn−1 . (i) If even(σ) = k − 1, then we can insert the two copies of n right after σ2i−1 , where i ∈ [n − 1]. (ii) If even(σ) = k, then we can insert the two copies of n in the front of σ or right after σ2i , where i ∈ [n − 1]. Clearly, E(1, 0) = 1 corresponds to 11 ∈ Q1 . Therefore, the numbers E(n, k) satisfy the recurrence relation E(n, k) = (n − 1)E(n − 1, k − 1) + nE(n − 1, k),

(6)

with the initial conditions E(1, 0) = 1 and E(1, k) = 0 for k ≥ 1. It follows from (6) that En (q) = (n + (n − 1)q)En−1 (q) for n ≥ 1, with the initial value E0 (q) = 1. Thus En (q) =

n Y

(i + (i − 1)q).

i=1

Recall that

n X n k=0

Therefore,

k

xk = x(x + 1) · · · (x + n − 1).

En (q) =

n X n k=0

which leads to (5), since

k

(1 + q)n−k ,

En (p, q) = pn En

q . p

6


3. Stirling derangements and increasing binary trees Let [k]n denote the set of words of length n in the alphabet [k]. For ω = ω1 ω2 · · · ωn ∈ [k]n , the reduction of ω, denoted by red(ω), is the unique word of length n obtained by replacing the ith smallest entry by i. For example, red(33224547) = 22113435. Very recently, Ma and Yeh [12] introduced the definition of Stirling permutations of the second kind. A permutation σ of the multiset [n]2 is a Stirling permutation of the second kind of order n whenever σ can be written as a nonempty disjoint union of its distinct cycles and σ has a standard cycle form that satisfies the following conditions: (i) For each i ∈ [n], the two copies of i appear in exactly one cycle; (ii) Each cycle is written with one of its smallest entry first and the cycles are written in increasing order of their smallest entries; (iii) The reduction of the word formed by all entries of each cycle is a Stirling permutation. In other words, if (c1 , c2 , . . . , c2k ) is a cycle of σ, then red(c1 c2 · · · c2k ) ∈ Qk . Let Q2n denote the set of Stirling permutations of the second kind of order n. In the following discussion, we always write σ ∈ Q2n in standard cycle form. Let (c1 , c2 , . . . , c2k ) be a cycle of σ, where k ≥ 2. An entry ci is called a cycle ascent plateau if ci−1 < ci = ci+1 , where 2 ≤ i ≤ 2k−1. Denote by cap(σ) (resp. cyc(σ)) the number of cycle ascent plateaux (resp. cycles) of σ. An entry k ∈ [n] be called a fixed point of σ if (kk) is a cycle of σ. Let fix(σ) denote the number of fixed points of σ. Using the fundamental transformation of Foata and Sch¨ utzenberger [6], we have X X q cyc(τ ) xcap(τ ) y fix(τ ) , (7) q lrmin(σ) xap(σ) y bk2 (σ) = τ ∈Q2n

σ∈Qn

where bk2 (σ) is the number of blocks of σ with length 2. The exponential generating function could be obtained as follows. Theorem 7. X X

q cyc(τ ) xcap(τ ) y fix(τ )

n≥0 τ ∈Q2n

q tn = A x, t et(y−1) , n!

where A(x, z) is the exponential generating function given by (1). Proof. In the same spirit as Theorem 5, the generating function is equal to X X

q cyc(π) xasc(π) y fix(π)

n≥0 π∈Sn

tn , n!

with the help of (7). The statistic fix(π) denotes the number of blocks of size 1 in π. We use the language of combinatorial objects as in the book of Flajolet and Sedgewick [5]. Take C to be the class of permutations in which the atom with label 1 is in the first place. Then Set(C) is exactly the class of permutations. Let the generating function of C be Gen(C; asc) := f (x, t) = t + x

t3 t2 + (x + x2 ) + O(t4 ), 2! 3!


7

where the variable x records the statistic ascent. The interpretation of Set(C) tells us that ef (x,t) = A(x, t). Moreover, X X

q cyc(π) xasc(π)

n≥0 π∈Sn

tn = eq·f (x,t) = A(x, t)q n!

is also straightforward. Now to take the statistic fix into account, the first-order term of f (x, t) should be changed from t to yt. Hence the generating function in question is q eq·(f (x,t)−t+yt) = A x, t et(y−1)

A perfect matching of [2n] is a set partition of [2n] with blocks (disjoint nonempty subsets) of size exactly 2. Let M2n be the set of matchings of [2n], and let M ∈ M2n . The standard form of M is a list of blocks {(i1 , j1 ), (i2 , j2 ), . . . , (in , jn )} such that ir < jr for all 1 ≤ r ≤ n and 1 = i1 < i2 < · · · < in . Throughout this paper we always write M in the standard form. If (is , js ) is such a block of M, then we say that is (resp. js ) is an ascending (resp. a descending) entry of M. Definition 8. Let hk be the number of perfect matchings of [4k] such that 2i − 1 and 2i are either both ascending or both descending for every i ∈ [2k]. Let ı2 = −1. Following [15], we have p

sec(2ız) =

X

(−1)n hn

n≥0

z 2n . (2n)!

Definition 9. A Stirling derangement is a Stirling permutation without blocks of length 2. Let DQn be the set of Stirling derangements of order n, i.e., DQn = {σ ∈ Qn | bk2 (σ) = 0}. The following result (in an equivalent form) has been algebraically obtained by Ma and Yeh [12, Page 15]. Theorem 10. X

σ∈DQn

(−1)ap(σ) =

(

0, if n is odd; k (−1) hk , if n = 2k is even.

In the rest of this section, we shall present a bijective proof of Theorem 10. Let ϕ be the bijection between permutations and increasing binary trees defined by the inorder traversal in depth-first search. The left-to-right minima of π ∈ Sn will be the labels in the leftmost path of the corresponding trees. Let Tn denote the set of bicolored increasing binary trees on n nodes such that all the nodes in the leftmost path are white and the other nodes are white or black. Then ϕ is a bijection from Sn to Tn if we match white nodes (resp. black nodes) with those unmarked letters (resp. marked letters.) Note that all of the left-to-right minima of a marked permutation are mapped to the nodes of the leftmost path of its increasing binary tree. These nodes will be called special nodes.

8


1

2

4

3

5

6

Figure 1. The bicolored tree corresponded to 2 3 1 5 4 6 For example, the Stirling permutation σ = 223315544166 corresponds to the marked permutation π = 2 3 1 5 4 6, which in turn corresponds to the bicolored increasing tree with 6 nodes in Figure 1. Note that the special nodes are labelled 1 and 2. A bijective proof of Theorem 10. Now we consider the composition ϕ ◦ Φ from Qn to Tn , where Φ is defined in the proof of Theorem 2. It is easily observed that k is an ascent plateau of σ ∈ Qn if and only if the node k in the bicolored increasing binary tree ϕ(θ(σ)) is non-special and does not have a left child. Also the Stirling derangements are mapped to those trees each of whose special nodes has a right child. For trees in Tn , there are some trees that have a non-special node with only one child. An involution can be introduced on these trees: among the non-special nodes with only one child, find the one with the minimal label, then move its whole subtree to the other branch. An example on this involution is shown in Figure 2. 1

1

2

3

4

5

6

7

2

3

4

5

6

7

8 π = 4 6 3 8 7 5 12

8 π b = 46358712

Figure 2. Involution on trees in Tn ; the move is made on Node 5.

Following the discussion in the previous paragraph, we immediately see that this involution is sign-reversing if we attach the sign (−1)ap(σ) to the tree ϕ(θ(σ)).


9

For those trees to which the above involution cannot be applied, all of its nodes have either zero or two children; these trees are called complete, whose number of nodes n must be even, and it has the sign (−1)ap(σ) = (−1)n/2 . From here we conclude that when n is odd, these trees are sign-balanced. Our remaining task is to give a proper count on those complete bicolored increasing binary trees on n = 2k nodes. Let σ ∈ DQn such that ϕ(θ(σ)) is complete. Then the intermediate marked permutation π = θ(σ) = π1 π2 . . . πn is reverse alternating, i.e., π1 < π2 > π3 < · · · . Note that in these situations the left-to-right minima of π always occur at odd-indexed positions. We now apply the Foata transformation on π by making its left-to-right minima as the head of each cycle; say π 7→ C1 C2 . . . Cj . Each cycle determines some pairings on ±[n]: for any C = (c1 c2 . . . cℓ ), we pair ci with ci+1 for 1 ≤ i < ℓ, and pair cℓ with c1 . Since the number at the beginning of a cycle must be unmarked, we have a bijection between complete bicolored increasing binary trees on n nodes with those special perfect matchings on [4k] = [2n] ≃ ±[n]. Using definition of the numbers hk , we get the desired result. Example 11. We use the ordering 1 < 1 < 2 < 2 < · · · < n < n on ±[n] ∼ = [2n] = [4k]. The reverse alternating marked permutation π = 3 5 1 6 2 4 corresponds to the following perfect matching on ±[6]: 3 5/5 3/1 6/6 2/2 4/4 1. Because π is reverse alternating, the numbers i and ı are both ascending or both descending for all i ∈ [n] in the corresponding perfect matching. 4. On r-colored marked permutations The definition of marked permutations can easily be extended to the r-colored version for any positive integer r. We think that a marked permutation (resp. an ordinary permutation) is in the special case r = 2 (resp. r = 1) in which the unmarked elements are painted by the first color, while the others are painted by the second color. The definition for the r-colored marked permutations is similar to that in the two-colored situation, i.e., every left-to-right minimum of π must be painted by the first color, which is always referred as white in this section. We find that on the r-colored marked permutations the ascent statistic is equidistributed with another statistic, which is defined below. (r) (r) Let Sn be the set of r-colored marked permutations of [n]. For π ∈ Sn , we define desrlmin(π) = {πi | πi−1 > πi for i > 1, or πi is a right-to-left minimum of π} . Theorem 12. Let r be a positive integer. The following polynomial is symmetric in the variables x and y: X xasc(π)+1 y desrlmin(π) . Fr (x, y) = (r)

π∈Sn

10


Proof. We will construct an involution ψ on Sn first. Let π = π1 π2 . . . πn ∈ Sn . Again we divide π into several blocks by marking each right-to-left minimum of π as the end of a block; we will view each block as a cycle. Now we rotate each block in order to list its largest number at the end of its block; then arrange these blocks so that these largest numbers are in decreasing order; lastly take the complement of every number to get the final result ψ(π). For example, let us go through step by step on the permutation π = 214853697 ∈ S9 : π = 214853697 → 21|4853|6|97 → 12|5348|6|79 → 79|5348|6|12 → 31|5762|4|98 7→ 315762498 = ψ(π). A careful examination shows that ψ is an involution on Sn , and asc(π) + 1 = desrlmin(ψ(π)). For r > 1, we can take a certain algorithm to ensure that every number in ψ(π) is still colored white. For example, consider the colors are taken modulo r by addition in each block; or during the block rotation process we keep the colors fixed at their respective positions. Since the above involution ψ does not depend on the coloring, we still have the result in symmetry. The generating function Fr (x, y) :=

X

xasc(π)+1 y desrlmin(π)

(r)

π∈Sn

can be obtained by using context-free grammars, as done by Dumont [4]. For r = 1, the grammar is {a → ab, b → ab, c → ab, e → ce}, which had been studied by Roselle [14]. Let A(x, y, z, t) be a refinement for the Eulerian polynomials: A(x, y, z, t) = zt +

X X

xasc(π)+1 y des(π)

n≥2 π∈Sn

tn . n!

Then the generating function for F1 (x, y) is shown in [4] to be G1 (x, y, t) :=

X

n≥0

We define Gr (x, y, t) :=

X

n≥0

Fr (x, y)

F1 (x, y)

tn = exp A(x, y, xy, t) . n!

tn . For any r-colored permutation, it can also be partitioned n!

into blocks by the method described in the proof of Theorem 12. Then we can assign those blocks whose end entries are painted by the same color into the same group. It is clear that the number of r-colored marked permutations of length n is the same as those r-colored permutations of length n whose right-to-left minima are all painted by any other color. Hence we have the identity: r G1 (x, y, rt) = Gr (x, y, t) ,

that is,

1/r Gr (x, y, t) = G1 (x, y, rt) .


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5. An interpretation of r-colored marked permutations Suppose a class of combinatorial objects C has the generating function Gen(C) = f (x, t), where x is the vector x = (x1 , x2 , . . . , xs ), and the power of xi records certain statistic stati on C. Here the generating function can be either ordinary or exponential, as an ordinary one can be seen as an exponential one with arbitrary labelling. Recall that g(x, t, r1 , r2 , . . . , rk ) := f (x, (r1 + r2 + · · · + rk )t) counts a weighted version of C, denoted by k C, where each atom in an object in C is weighted by a number in [k]. An object in k C with n atoms could be seen as (C, w), where C ∈ C and w is a map from [n] to [k]. Then the power of xi in g records stati (C), and the power of rj records the number of atoms in C with weight j, i.e., the cardinality of the pre-image w−1 (j). It is well-known that for a given non-negative integer k, fseq,k (x, t) := f (x, t)k counts Seqk (C). Each object in Seqk (C) is of the form of a k-sequence: (C1 , C2 , . . . , Ck ) with each Ci ∈ C. The power of xi in hs records stati (C1 ) + stati (C2 ) + · · · + stati (Ck ). On the other hand, f (x, t)k fset,k (x, t) := k! counts Setk (C). This induced class contains objects of the form of a k-set {C1 , C2 , . . . , Ck } with each Ci ∈ C. The power of xi in hs also records stati (C1 ) + stati (C2 ) + · · · + stati (Ck ). When C has no objects of size 0, the power series 1 fseq (x, t) := 1 − f (x, t) is well-defined, and it counts Seq(C) := ∪∞ k=0 Seqk (C); while the power series fset (x, t) := exp (f (x, t)) counts Set(C) := ∪∞ k=0 Setk (C). Suppose now that C is a combinatorial objects with no objects of size 0, with generating function Gen(C) = f (x, t). Observe the formal power series 1 f (x, kt) k has non-negative integer coefficients. This fact suggests that there should be some general method to induce a new class ∗k C of combinatorial objects that is enumerated by h. Suppose we have a rule to specify an atom from any object C in C; that atom will be called a distinguished atom of C. For example, let the distinguished atom be the one with label 1. Then we could define ∗k C to be a weighted version of C, where each atom except the distinguished one in an object in C is weighted by a number in [k]. The combinatorial interpretation of the refined generating function 1 f (x, (r1 + r2 + · · · + rk )t) h(x, t, r1 , r2 , . . . , rk ) := r1 + r2 + · · · + rk h(x, t) :=

is straightforward.

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Take C to be the class of permutations in which the atom with label 1 is in the first place. Then Set(C) (resp. Set(∗r C)) is exactly the class of permutations (resp. r-colored marked permutations) that we introduced in this article. References [1] M. B´ ona. Real zeros and normal distribution for statistics on Stirling permutations defined by Gessel and Stanley. SIAM J. Discrete Math., 23:401–406, 2008/09. [2] F. Brenti. A class of q-symmetric functions arising from plethysm. J. Combin. Theory Ser. A, 91:137–170, 2000. [3] W.Y.C. Chen, A.M. Fu. Context-free grammars for permutations and increasing trees, Adv. in Appl. Math., 82:58–82, 2017. [4] Dominique Dumont. Grammaires de William Chen et dérivations dans les arbres et arborescences, Sém. Lothar. Combin., 37 (1996), Art. B37a–21 pp. (electronic). [5] Philippe Flajolet and Robert Sedgewick. Analytic combinatorics. Cambridge University Press, Cambridge, 2009. [6] D. Foata, M. Sch¨ utzenberger. Théorie Géométrique des Polynˆ omes Euleriens. Lecture Notes in Mathematics, vol. 138, Springer-Verlag, Berlin-New York, 1970. [7] I. Gessel and R.P. Stanley. Stirling polynomials. J. Combin. Theory Ser. A, 24:25–33, 1978. [8] J. Haglund, M. Visontai. Stable multivariate Eulerian polynomials and generalized Stirling permutations, European J. Combin., 33:477–487, 2012. [9] S. Janson, M. Kuba and A. Panholzer. Generalized Stirling permutations, families of increasing trees and urn models, J. Combin. Theory Ser. A, 118:94–114, 2011. [10] S.-M. Ma, T. Mansour. The 1/k-Eulerian polynomials and k-Stirling permutations. Discrete Math., 338:1468– 1472, 2015. [11] S.-M. Ma, Y.-N. Yeh. Stirling permutations, cycle structure of permutations and perfect matchings. Electron. J. Combin., 22(4):#P4.42, 2015. [12] S.-M. Ma, Y.-N. Yeh. Eulerian polynomials, perfect matchings and Stirling permutations of the second kind. arXiv:1607.01311. [13] J.B. Remmel, A.T. Wilson. Block patterns in Stirling permutations. arXiv:1402.3358. [14] D P Roselle. Permutations by number of rises and successions, Proc. Amer. Math. Soc., 19(1): 8–16, 1968. [15] C.V. Sukumar and A. Hodges. Quantum algebras and parity-dependent spectra. Proc. R. Soc. A 463: 2415– 2427, 2007. Institute of Mathematics, Academia Sinica, Taipei, Taiwan E-mail address: [email protected] (G.-H. Duh) Department of Mathematics, National Taiwan Normal University, Taipei 116, Taiwan E-mail address: [email protected] (Y.-C. Lin) School of Mathematics and Statistics, Northeastern University at Qinhuangdao, Hebei 066004, P.R. China E-mail address: [email protected] (S.-M. Ma) Institute of Mathematics, Academia Sinica, Taipei, Taiwan E-mail address: [email protected] (Y.-N. Yeh)